EEC the mission

[Login to see usernames]

In this example the amount of work done by the person is 30J, this is also the amount of electrical potential energy that is possessed by all three charges together.  The electrical potential (not energy) is the amount of energy per unit of charge. 
 
(http://regentsprep.org/Regents/physics/phys03/apotdif/calc.gif)

At the original position of the charges they have no energy, so they also have no electrical potential or 0 volts.  Once they are pulled apart, they have have an electrical potential of 10 volts.  We could say that the electrical potential difference from one point to the other is 10 volts.

Keep in mind that the electrical potential describes the amount of energy per unit of charge.  This means that when one of the charges is released, the electric field will do 10 Joules of work on the charge so it will have a kinetic energy of 10 Joules the instant before it strikes the negative charge.

[Login to see usernames]

Earlier in this thread i posted that may have reduced my credibility hahaha

 stated that the farther the ions are from each other the greater is the force of attraction ... well i don't know where i was with my head that night

actually the farther they are the weaker is the field and it decreases with the square of the distance... so Once the ions get separated from each other it won't take too much work to separate them further...

is it correct now?

F=k*q*Q/dˆ2

E=F/q= Q*k/dˆ2

I think i was confusing with electric potential and sincerely i way too tired these times...

I was soo confused......

[Login to see usernames]

F=k*q*Q/dˆ2 if the result is positive the force is repelling if negative (opposite charges) its attractive force

E=F/q= Q*k/dˆ2   This is electric field (vector) of a point charge so the farther from the point charge the weaker is the field..

U=kqQ/d     this is the potential energy if the charges are positive their potential is greater when they are the closest together such that the potential at infinite is zero....

V= (kQ/d1)-(kQ/d2)    this is the electric potential if d2 is set to infinite the right side turns to zero and so the V=kQ/d

If there are two charges you need both to know the potential energy between them and it can be summed because its a scalar quantity

k=1/4*pi*Epsilon0

if C=Q/v than C=Q/Q/4*pi*epsilon*d=  4*pi*epsilon*r (self capacitance of a point charge or sphere where the r is the radius)

[Login to see usernames]

the 14 equation at meyer tech brief is wrong

the voltage potential due to a charge q at a distance d where the dielectric constant of the medium is Er actually is V=q/4*Pi*epsilon*Er*d


I think he shows this equation to let us conclude that the electric potential reduces by the dielectric constant factor

of course if there are two point charges at a distance d than the potential will double...


is all this correct?

[Login to see usernames]

C=Q/V , dielectrics are also used to increase capacitance by reducing voltage *.  Also electric flow Φ=int(E*ds)=q/ε0 (gaussian surfaces).  Potential is a scalar and electric field a vector and if there are more point charges V=k(q1/r1+q2/r2) so yes all correct.

voltage is decreased because electric field is decreased*